Difference between revisions of "2003 Pan African MO Problems/Problem 3"
Rockmanex3 (talk | contribs) (Solution to Problem 3 -- easy number base problem) |
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&= (b^2 + b + 1)(b^2 - b + 1) | &= (b^2 + b + 1)(b^2 - b + 1) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | In order for <math>(b^2 + b + 1)(b^2 - b + 1)</math> to be prime, either <math>b^2 + b + 1</math> or <math>b^2 - b + 1</math> (but not both) must equal <math>1</math>. If <math>b^2 + b + 1 = 1</math>, then <math>b = 0</math> or <math>b = -1</math>, but none of the values of <math>b</math> are valid base numbers. If <math>b^2 - b + 1 = 1</math>, then <math>b = 0</math> or <math>b = 1</math>. However, neither value are valid bases because <math>0</math> is less than <math>1</math> and the number <math>10101</math> has | + | In order for <math>(b^2 + b + 1)(b^2 - b + 1)</math> to be prime, either <math>b^2 + b + 1</math> or <math>b^2 - b + 1</math> (but not both) must equal <math>1</math>. If <math>b^2 + b + 1 = 1</math>, then <math>b = 0</math> or <math>b = -1</math>, but none of the values of <math>b</math> are valid base numbers. If <math>b^2 - b + 1 = 1</math>, then <math>b = 0</math> or <math>b = 1</math>. However, neither value are valid bases because <math>0</math> is less than <math>1</math> and the number <math>10101</math> has ones for digits (making base <math>1</math> an invalid base). |
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Latest revision as of 02:44, 29 July 2023
Problem
Does there exists a base in which the numbers of the form:
are all prime numbers?
Solution
Using the definition of base numbers, the number in base
can be rewritten as
, where
.
The above expression can be factored, so
In order for
to be prime, either
or
(but not both) must equal
. If
, then
or
, but none of the values of
are valid base numbers. If
, then
or
. However, neither value are valid bases because
is less than
and the number
has ones for digits (making base
an invalid base).
Therefore, there is no base where are all prime numbers.
See Also
2003 Pan African MO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All Pan African MO Problems and Solutions |